In fact, the above theorem is really another version of the completeness axiom. Thus its
validity implies completeness. You might try to show this.

One of the interesting things about absolutely convergent series is that you can “add them
up” in any order and you will always get the same thing. This is the meaning of
the following theorem. Of course there is no problem when you are dealing with
finite sums thanks to the commutative law of addition. However, when you have
infinite sums strange and wonderful things can happen because these involve a limit.

Theorem 5.1.10Let θ : ℕ → ℕbe one to one and onto. Suppose

∑∞
ak
k=1

converges absolutely. Then

∞∑ ∑∞
aθ(k) = ak
k=1 k=1

Proof:From absolute convergence, there exists M such that

∑∞
|ak| < ε
k=M+1

Since θ is one to one and onto, there exists N ≥ M such that

{1,2,⋅⋅⋅,M }

⊆

{θ(1),θ(2),⋅⋅⋅,θ (N )}

.
It follows that it is also the case that

∑∞ || ||
aθ(k) < ε
k=N+1

This is because the partial sums of the above series are each dominated by a partial sum for
∑k=M+1∞

|ak|

since every index θ

(k)

equals some n for n ≥ M + 1. Then since ε is
arbitrary, this shows that the partial sums of ∑aθ

Thus the even partial sums are decreasing and the odd partial sums are increasing. The
even partial sums are bounded below also. (Why?) Therefore, the limit of the even
partial sums exists. However, it must be the same as the limit of the odd partial
sums because of the last equality above. Thus limn→∞Sn exists and so the series
converges. Now I will show later below that ∑k

21k

and ∑k

21k−1

both diverge.
Include enough even terms for the sum to exceed 7. Next add in enough odd terms so
that the result will be less than 7. Next add enough even terms to exceed 7 and
continue doing this. Since 1∕k converges to 0, this rearrangement of the series must
converge to 7. Of course you could also have picked 5 or −8 just as well. In fact,
given any number, there is a rearrangement of this series which converges to this
number.

Theorem 5.1.12(comparison test) Suppose

{an}

and

{bn}

are sequencesofnon negative real numbers and suppose for all n sufficiently large, an≤ bn. Then

If∑n=k∞bnconverges, then∑n=m∞anconverges.

If∑n=k∞andiverges, then∑n=m∞bndiverges.

Proof: Consider the first claim. From the assumption, there exists n∗ such that
n∗> max

(k,m)

and for all n ≥ n∗bn≥ an. Then if p ≥ n∗,

∗
∑p n∑ ∑k
an ≤ an + ∗ bn
n=m n=m∗ n=n+1
n∑ ∞∑
≤ an + bn.
n=m n=k

Thus the sequence,

∑p
{ n=m an}

p=m∞ is bounded above and increasing. Therefore, it
converges by completeness. The second claim is left as an exercise. ■

Example 5.1.13Determine the convergence of∑n=1∞

1-
n2

.

For n > 1,

1 1
n2 ≤ n-(n−-1).

Now

∑p 1 ∑p [ 1 1]
n(n-−-1) = n-−-1 − n-
n=2 n=2
= 1− 1 → 1
p

Therefore, letting an =

n12

and bn =

n(1n−1)

A convenient way to implement the comparison test is to use the limit comparison test.
This is considered next.

Theorem 5.1.14Let an,bn> 0 and suppose for all n large enough,

an- an-
0 < a < bn ≤ bn < b < ∞.

Then∑anand∑bnconverge or diverge together.

Proof:Let n∗ be such that n ≥ n∗, then

an an
---> a and---< b
bn bn

and so for all such n,

ab < a < bb
n n n

and so the conclusion follows from the comparison test. ■

The following corollary follows right away from the definition of the limit.

Corollary 5.1.15Let an,bn> 0 and suppose

an-
lnim→∞ bn = λ ∈ (0,∞ ).

Then∑anand∑bnconverge or diverge together.

Example 5.1.16Determine the convergence of∑k=1∞

---1----
√n4+2n+7

.

This series converges by the limit comparison test above. Compare with the series of
Example 5.1.13.

Therefore, the series converges with the series of Example 5.1.13. How did I know what to
compare with? I noticed that

----------
√ n4 + 2n +7

is essentially like

√ --
n4

= n2 for large enough n.
You see, the higher order term n4 dominates the other terms in n4 + 2n + 7. Therefore,
reasoning that 1∕

----------
√n4 + 2n+ 7

is a lot like 1∕n2 for large n, it was easy to see what to
compare with. Of course this is not always easy and there is room for acquiring skill through
practice.

To really exploit this limit comparison test, it is desirable to get lots of examples of series,
some which converge and some which do not. The tool for obtaining these examples
here will be the following wonderful theorem known as the Cauchy condensation
test.

Theorem 5.1.17Let an≥ 0 and suppose the terms of the sequence

{an}

aredecreasing. Thus an≥ an+1for all n. Then

∑∞ ∑∞
an and 2na2n
n=1 n=0

converge or diverge together.

Proof:This follows from the inequality of the following claim.

Claim:

∑n ∑2n n∑
2ka2k−1 ≥ ak ≥ 2k− 1a2k.
k=1 k=1 k=0

Proof of the Claim: Note the claim is true for n = 1. Suppose the claim is true for n.
Then, since 2n+1− 2n = 2n, and the terms, an, are decreasing,

converge or diverge together. If p > 1, the last series above is a geometric series having
common ratio less than 1 and so it converges. If p ≤ 1, it is still a geometric series but in this
case the common ratio is either 1 or greater than 1 so the series diverges. It follows that the p
series converges if p > 1 and diverges if p ≤ 1. In particular, ∑n=1∞n−1 diverges while
∑n=1∞n−2 converges.

Example 5.1.19Determine the convergence of∑k=1∞

√n2+1100n-

.

Use the limit comparison test.

(1)
lnim→∞ (----n1---)-= 1
√n2+100n-

and so this series diverges with ∑k=1∞

1k

.

Sometimes it is good to be able to say a series does not converge. The nth term test
gives such a condition which is sufficient for this. It is really a corollary of Theorem
5.1.7.

It is very important to observe that this theorem goes only in one direction. That is, you
cannot conclude the series converges if limn→∞an = 0. If this happens, you don’t know
anything from this information. Recall limn→∞n−1 = 0 but ∑n=1∞n−1 diverges. The
following picture is descriptive of the situation.